Question

The molar heat of vaporization for water is 10.79 kJ/mol. How much energy must be absorbed by 100 grams
of water at 100 °C in order to convert it to steam at 100 °C?
A 4079 0
B 140.79 k)
C 7.35 k)
D 226 kJ

Answer 1

Answer:

59.878 Kilo joules

Explanation:

We are given;

• Molar heat of vaporization for water as 10.79 kJ/mol
• Mass of water 100 grams

We are required to calculate the amount of energy absorbed to convert water to ice.

• We need to know that the process of conversion of water to steam without change in temperature is known as vaporization or boiling.
• Therefore, heat absorbed is calculated by multiplying the mass of water or the number of moles by heat of vaporization.

• That is, Q = m × ΔHf
• When given the number of moles and the molar heat of vaporization, then Q = n × ΔHf

In this case;

Mass of water = 100 g

But, molar mass of water = 18.02 g/mol

Moles of water = 100 g ÷ 18.02 g/mol

                         = 5.549 Moles

Therefore;

Q = 5.549 moles × 10.79 kJ/mol

   = 59.878 Kilo joules